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Question: (i) The boiling point of benzene is 353.23K when 1.80 gram of non volatile solute was dissolved in 9...

(i) The boiling point of benzene is 353.23K when 1.80 gram of non volatile solute was dissolved in 90 gram of benzene. The boiling point is raised to 354.11K. Calculate the molar mass of the solute. [Kb{{K}_{b}} for benzene = 2.53Kmol12.53Kmo{{l}^{-1}}]
(ii) State Henry’s law

Explanation

Solution

Hint The answer for this question is based on the physical chemistry concept which tells about the elevation in the boiling point and the formula is given by ΔTb=Kbm\Delta {{T}_{b}}={{K}_{b}}m and the Henry’s law is applied to the gases at constant temperature.

Complete step – by – step answer:
In the lower classes of chemistry, we have studied about the physical chemistry concept which tells about the elevation in the boiling point, depression in the freezing point and relating terms and also about the laws of gas molecules like Henry’s law, Gay-Lussac’s law etc.
We shall now see the calculation of molar mass of the solute based on the elevation in the boiling point and also define what Henry’s law is.
(i) Elevation in boiling point is denoted by ΔTb\Delta {{T}_{b}} is defined as the phenomenon in which the boiling point of a liquid that is solvent will be higher when another compound that is non volatile solute is added to it. In simple words it means that solution has higher boiling point than that of pure solvent.
Now, in the above given data, the boiling point of benzene that is pure solvent is353.23K and there is an addition of non volatile solute where the boiling point rises to354.11K.
Thus, the difference in these two values gives the value ofΔTb\Delta {{T}_{b}}
ΔTb=354.11353.23=0.88K\Rightarrow \Delta {{T}_{b}}=354.11-353.23=0.88K
Now, since the formula for ΔTb\Delta {{T}_{b}}isΔTb=Kbm\Delta {{T}_{b}}={{K}_{b}}m
Substituting the values as per data given, we have
0.88=2.53×mole90×10000.88=2.53\times \dfrac{mole}{90}\times 1000
moles=0.0313\Rightarrow moles=0.0313
Since the weight of non volatile solute added is 1.8g, the molar mass can be calculated as,
1.8m=0.0313\dfrac{1.8}{m}=0.0313
m=1.80.0313=57.8g\Rightarrow m=\dfrac{1.8}{0.0313}=57.8g
Therefore, the correct answer is molar mass of solute = 57.8g
(ii) Henry’s law is given for gases at room temperature and it states that ‘At constant temperature, the amount of given gas that dissolves in a given type and volume of liquid is directly proportional to the partial pressure of that particular gas in equilibrium with that liquid’.

Note: Note that elevation in boiling point is also called as the ebullioscopy and the instrument used to measure this is called an ebullioscope where the molecular weight of a non volatile solute is determined based on the boiling point elevation.